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[Question forwarded from SE for lack of interaction]

Given an geometric genus $1$ curve with $1$ node $C$ (hence with arithmetic genus $2$), its normalization is given by the elliptic curve $\tilde{C}$, which is isomorphic as algebraic groups to its Jacobian, also denoted by $\tilde{C}$

I am trying to use the Leray-Serre spectral sequence associated to the following exact sequence of algebraic groups

\begin{equation} 1\to \mathbb{C}^*\to J_C \to \tilde{C}\to 0 \end{equation}

to compute the equivariant cohomology of the total space $J_C$.

Here by equivariant I mean the following: inversion of the two points of the node induces a multiplication by $(-1)$ on the fibre $\mathbb{C}^* $, which in turn, makes the cohomology group $H^1_c(\mathbb{C}^*,\mathbb{Q})\cong \mathbb{Q}s_{1^2}$ the sign representation of the permutation group $S_2$.

What I would like to know is a natural way to interpret such an action on $\tilde{C}$, i.e. how to compute $ H_c^*(\tilde{C},\mathbb{Q}s_{1^2}) $, which via the spectral sequence mentioned above would give the rational cohomology with compact support of $J_C$.

My naive attempt involved using group cohomology, since the $\tilde{C}$ is the classifying space of $\mathbb{Z}\oplus\mathbb{Z}$. The problem I am encountering is that I don't see how is $\mathbb{Q}s_{1^2}$ is a $\mathbb{Z}\oplus\mathbb{Z}$-module.

Any reccomendation on how to approach the problem, even with other methods is greatly appreciated.

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    $\begingroup$ If your curve has normalization an elliptic curve, then generally it would be referred to as having genus $2$, no? Since it is parametrized by a point in the boundary of $\overline{M}_2$. It's a small quibble but also a point of potential confusion. $\endgroup$ Commented Apr 12, 2023 at 16:14
  • $\begingroup$ @TabesBridges yes you are right, I meant arithmetic genus $1$ with $1$ node, hence geometric genus $2$. I'll make it clearer thank you. $\endgroup$
    – Marco Fava
    Commented Apr 12, 2023 at 16:28
  • $\begingroup$ I'm sorry. in my previous comment I have clearly swapped the terms "geometric" and "arithmetic". $\endgroup$
    – Marco Fava
    Commented Apr 13, 2023 at 15:51

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